Question

Prove the following statement: ${{\sec }^{2}}A{{\csc }^{2}}A={{\tan }^{2}}A+{{\cot }^{2}}A+2$

Answer 1

Hint: We are going to prove the above statement by using some trigonometric formulas and then we will start from RHS and then by using the formulas we will show that it is equal to LHS and just to be sure that the answer we have calculated is correct we will try to put some values in place of A to check if it’s matches or not.

Complete step-by-step answer:

Let’s start by solving RHS,

${{\tan }^{2}}A+{{\cot }^{2}}A+2$

Now we will use $\cot A=\dfrac{1}{\tan A}$ we get,

$\begin{align}


  & {{\tan }^{2}}A+{{\left( \dfrac{1}{\tan A} \right)}^{2}}+2 \\


 & =\dfrac{{{\tan }^{4}}A+2{{\tan }^{2}}A+1}{{{\tan }^{2}}A} \\


\end{align}$


Now we use,


${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$


Then RHS becomes,


$=\dfrac{{{\left( {{\tan }^{2}}A+1 \right)}^{2}}}{{{\tan }^{2}}A}$


Now we know that $1+{{\tan }^{2}}x={{\sec }^{2}}x$ using this we get,


We will also use these formula,


$\begin{align}


  & \tan A=\dfrac{\sin A}{\cos A} \\


 & \sec A=\dfrac{1}{\cos A} \\


 & \csc A=\dfrac{1}{\sin A} \\


\end{align}$


Using the above formula we get,


$\begin{align}


  & =\dfrac{{{\left( {{\sec }^{2}}A \right)}^{2}}}{{{\tan }^{2}}A} \\


 & =\dfrac{\dfrac{{{\sec }^{2}}A}{{{\cos }^{2}}A}}{\dfrac{{{\sin }^{2}}A}{{{\cos }^{2}}A}} \\


 & =\dfrac{{{\sec }^{2}}A}{{{\sin }^{2}}A} \\


 & ={{\sec }^{2}}A{{\csc }^{2}}A \\


\end{align}$

And hence we have shown that LHS = RHS by using the above formula.


Hence Proved.

Note: It’s always better that we check if the answer that we have got by using the above formula we is correct or not to avoid some calculation mistake and for that we need to put some values in place of A to check whether it satisfies the above expression or not, so let’s check by putting A = $\dfrac{\pi }{4}$, we get LHS = ${{\sec }^{2}}\dfrac{\pi }{4}{{\csc }^{2}}\dfrac{\pi }{4}={{\left( \sqrt{2} \right)}^{2}}{{\left( \sqrt{2} \right)}^{2}}=4$ , and now if we look at RHS we get RHS = ${{\tan }^{2}}\dfrac{\pi }{4}+{{\cot }^{2}}\dfrac{\pi }{4}+2=1+1+2=4$ , and hence LHS = RHS, hence from this we can easily say that the answer that we have calculated is correct and there is no calculation mistake. There are many ways to solve this question as one can start from LHS and then show that RHS = LHS.